How can heat change be measured under constant pressure?

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SUMMARY

Enthalpy is defined as the heat content of a system at constant pressure, represented by the equation H = U + PV. The enthalpy change (ΔH) is equal to the heat absorbed or evolved by the system when pressure remains constant, while internal energy change (ΔU) is equal to the heat absorbed or evolved at constant volume. These two changes differ due to the work done by the system during expansion, which is not accounted for in internal energy. Thus, enthalpy change is measured in terms of heat exchanged under constant pressure, independent of volume changes.

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TheExibo
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So enthalpy is the heat content of a system at constant pressure. Enthalpy change is equal to the heat absorbed or evolved by the system at constant pressure. If my understanding is correct, a system whose temperature goes up will return back to that starting temperature if pressure is kept constant (i.e., its volume is allowed to expand). Therefore, is enthalpy, or heat change, measured in terms of the expansion?

Additionally, would enthalpy change and internal energy change (which is equal to the heat absorbed or evolved by the system at constant volume) not have the same values? After all, in the former, the amount of energy released (if the reaction inside releases energy) can be calculated by multiplying the change in volume with the outside pressure, and this value of energy, as far as I know, should be equal to the energy of the change in temperature inside the latter system.
 
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Heat is not a state variable, i.e., it is not a property of a system, it is the energy that is thermally exchanged between two systems. This follows from the fact that ##\delta Q## is not an exact differential.
 
Enthalpy change is not $\Delta U+P\Delta V$. It is $$\Delta U+\Delta (PV)$$. So, even at constant volume, if the pressure changes, the enthalpy change is not equal to the internal energy change.
 
The relationship between enthalpy ##H## and internal energy ##U## is ##H=U+PV##.
In a reversible process we have ##dU=dQ+dW=dQ-PdV##, and we have ##dH=dQ+VdP##.
So ##\Delta H = \Delta Q + \int_{P_1}^{P_2} V dP##. If the pressure is indeed constant, this becomes ##\Delta H=\Delta Q##, which is then independent of volume.
Similarly ##\Delta U = \Delta Q - \int_{V_1}^{V_2} PdV##. If the volume is constant, this becomes ##\Delta U=\Delta Q##, which is independent of pressure.
In both cases the change in enthalpy respectively in internal energy is simply the heat that is applied to the system. Keep in mind though that these are different processes with different final states.
 
Last edited:
TheExibo said:
So enthalpy is the heat content of a system at constant pressure.

No, it isn't. Enthalpy is fedined as

H = U + p \cdot V

TheExibo said:
Enthalpy change is equal to the heat absorbed or evolved by the system at constant pressure.

Yes that's correct and the reason for the definition of enthalpy. Calorimetirc measurements of internal energy would require constant volume which is much harder to achieve.

TheExibo said:
If my understanding is correct, a system whose temperature goes up will return back to that starting temperature if pressure is kept constant (i.e., its volume is allowed to expand).

I'm afraid your understanding is not correct.

TheExibo said:
Therefore, is enthalpy, or heat change, measured in terms of the expansion?

Change of enthalpy is measured in terms of exchanged heat under constant pressure. Expansion is just a possible side effect of the constant pressure and irrelevant for the measurement.

TheExibo said:
Additionally, would enthalpy change and internal energy change (which is equal to the heat absorbed or evolved by the system at constant volume) not have the same values?

No. Under constant pressure they differ by the volumetric work.
 

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